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Polydivisible number
In mathematics a polydivisible number (or magic number) is a number with digits abcdef... that has the following properties : # Its first digit a'' is not 0. # The number formed by its first two digits ''ab is a multiple of 2. # The number formed by its first three digits abc is a multiple of 3. # The number formed by its first four digits abcd is a multiple of 4. # etc. For example, EX9876 is a six-digit polydivisible number, but 123456 is not, because 12345 is not a multiple of 5. Polydivisible numbers can be defined in any base - however, the numbers in this article are all in base 10, so permitted digits are 0 to E. There are 71823 polydivisible numbers, and the largest of them is 24-digit 606890346850EX6800E036206464 If k'' is a polydivisible number with ''n-1 digits, then it can be extended to create a polydivisible number with n'' digits if there is a number between 10''k and 10''k''+E that is divisible by n''. If ''n is less or equal to 10, then it is always possible to extend an (n''-1)-digit polydivisible number to an ''n-digit polydivisible number in this way, and indeed there may be more than one possible extension. If n'' is greater than 10, it is not always possible to extend a polydivisible number in this way, and as ''n becomes larger, the chances of being able to extend a given polydivisible number become smaller (e.g. for n''=14, the chance is 3/4 or 90%, and for ''n=16, the chance is 2/3 or 80%, while for n''=20, the chance is only 1/2 or 60%). On average, each polydivisible number with ''n-1 digits can be extended to a polydivisible number with n'' digits in 10/''n different ways. This leads to the following estimate for F(n) : F''(''n) ≈ (E*10^(n''-1))/(''n!) Summing over all values of n'', this estimate suggests that the total number of polydivisible numbers will be approximately E*(''e^10-1)/10 ≈ 72407 where e'' = 2.875236069822... is the base of natural logarithm. There are about E*10^(''n-1)/''n''! n''-digit polydivisible numbers. There are no E-digit polydivisible numbers using all the digits 1 to E exactly once. (hence there are also no 10-digit polydivisible numbers using all the digits 0 to E exactly once, since if a number with digits ''abcdefghijkl is a 10-digit polydivisible number using all the digits 0 to E exactly once, then {a'', ''b, c'', ''d, e'', ''f, g'', ''h, i'', ''j, k'', ''l} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, and then abcdefghijkl is divisible by 10, thus we have l'' = 0 (by divisibility rule of 10), and {''a, b'', ''c, d'', ''e, f'', ''g, h'', ''i, j'', ''k} = {1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, thus a number with digits abcdefghijk is an E-digit polydivisible numbers using all the digits 1 to E exactly once). (proof: if a number with digits abcdefghijk is an E-digit polydivisible numbers using all the digits 1 to E exactly once, then {a'', ''b, c'', ''d, e'', ''f, g'', ''h, i'', ''j, k''} = {1, 2, 3, 4, 5, 6, 7, 8, 9, X, E}, and we have: ''f = 6 (since abcdef is divisible by 6) (by divisibility rule of 6) {d'', ''h} = {4, 8} (since abcd is divisible by 4 and abcdefgh is divisible by 8 (thus by 4)) (by divisibility rule of 4) {c'', ''i} = {3, 9} (since abc is divisible by 3 and abcdefghi is divisible by 9 (thus by 3)) (by divisibility rule of 3) {b'', ''j} = {2, X} (since ab is divisible by 2 and abcdefghij is divisible by X (thus by 2)) (by divisibility rule of 2) thus, we have {a'', ''e, g'', ''k} = {1, 5, 7, E} Since abcdefgh is divisible by 8, thus gh is divisible by 8 (by divisibility rule of 8), and since {a'', ''e, g'', ''k} = {1, 5, 7, E}, thus g'' is odd, and ''h must be 4 (if h'' = 8 and ''g is odd, then gh is not divisible by 8), and since abcdefghi is divisible by 9, thus hi is divisible by 9 (by divisibility rule of 9), however, h'' = 4 and ''i is either 3 or 9, but neither 43 nor 49 is divisible by 9. If we do not require the number formed by its first 8 digits divisible by 8, then there are 2 solutions: 1X98265E347 and 7298X65E341 (neither satisfies that the number formed by its first 8 digits is divisible by 4). If we do not require the number formed by its first 9 digits divisible by 9, then there are 4 solutions: 1X38E694725, 7X981634E25, 7X98E654321, and EX987634125 (only 7X98E654321 satisfies that the number formed by its first 9 digits is divisible by 3). Except this case of dozenal, such number exists in all even bases < 14 (and does not exist in any odd base since the number formed by the first (base−1) digits cannot be divisible by (base−1)), however, such number also does not exist in any even base 14 ≤ b ≤ 48. Category:Pages